Solve the system by the method of your choice. Identify systems with no solution and systems with infinitely many
solutions, using set notation to express their solution sets.
-$$\begin{array} { l }
\frac { x } { 3 } + \frac { y } { 6 } = 1 \
\frac { x } { 2 } - \frac { y } { 4 } = 0
\end{array}$$
A) $\left\{ \left( \frac { 3 } { 2 } , 3 \right) \right\}$
B) $\left\{ \left( 3 , \frac { 3 } { 2 } \right) \right\}$
C) $\left\{ ( x , y ) \mid \frac { x } { 3 } + \frac { y } { 6 } = 1 \right\}$
D) $\varnothing$

Question

Solve the system by the method of your choice. Identify systems with no solution and systems with infinitely many
solutions, using set notation to express their solution sets.
-$$\begin{array} { l } 
\frac { x } { 3 } + \frac { y } { 6 } = 1 \
\frac { x } { 2 } - \frac { y } { 4 } = 0
\end{array}$$
A) $\left\{ \left( \frac { 3 } { 2 } , 3 \right) \right\}$
B) $\left\{ \left( 3 , \frac { 3 } { 2 } \right) \right\}$
C) $\left\{ ( x , y ) \mid \frac { x } { 3 } + \frac { y } { 6 } = 1 \right\}$
D) $\varnothing$

Quizplus · Accepted Answer

To solve the system by elimination method, we can rewrite the two equations by multiplying the first equation by 2 and the second equation by 3:
$$\begin{cases} 2\left(\dfrac{x}{3}+\dfrac{y}{6}\right) = 2\cdot 1 \ 3\left(\dfrac{x}{2}-\dfrac{y}{4}\right) = 3\cdot 0\end{cases}$$
$$\begin{cases} \dfrac{x}{3}+\dfrac{y}{3} = 2 \ \dfrac{3x}{2}-\dfrac{3y}{4} = 0\end{cases}$$
We can see that the second equation is equivalent to the first equation, so it is redundant. Therefore, we can just use the first equation to solve for $x$ and $y$:
$$\frac{x}{3} + \frac{y}{3} = 2 \quad \Longrightarrow \quad x+y=6 \quad \Longrightarrow \quad y = 6-x$$
Substituting $y=6-x$ into the first equation, we get:
$$\frac{x}{3} + \frac{6-x}{6} = 1 \quad \Longrightarrow \quad \frac{x}{2} = \frac{1}{2} \quad \Longrightarrow \quad x=3$$
Therefore, $y=6-x=3$, and the solution is $(x,y)=(3,3)$. Checking the solution in the original system, we can see that it satisfies both equations.

Thus, the solution set is $\{(3,3)\}$, which corresponds to choice A.

Solve the System by the Method of Your Choice x3+y6=1x2−y4=0\begin{array} { l } \frac { x } { 3 } + \frac { y } { 6 } = 1 \\\frac { x } { 2 } - \frac { y } { 4 } = 0\end{array}3x​+6y​=12x​−4y​=0​

Solve the System by the Method of Your Choice $\begin{array} { l } \frac { x } { 3 } + \frac { y } { 6 } = 1 \\\frac { x } { 2 } - \frac { y } { 4 } = 0\end{array}$